733. Flood Fill

Description

An image is represented by an m x n integer grid image where image[i][j] represents the pixel value of the image.

You are also given three integers sr, sc, and color. You should perform a flood fill on the image starting from the pixel image[sr][sc].

To perform a flood fill, consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color), and so on. Replace the color of all of the aforementioned pixels with color.

Return the modified image after performing the flood fill.

Example 1:

Input: image = [[1,1,1],[1,1,0],[1,0,1]], sr = 1, sc = 1, color = 2
Output: [[2,2,2],[2,2,0],[2,0,1]]
Explanation: From the center of the image with position (sr, sc) = (1, 1) (i.e., the red pixel), all pixels connected by a path of the same color as the starting pixel (i.e., the blue pixels) are colored with the new color.
Note the bottom corner is not colored 2, because it is not 4-directionally connected to the starting pixel.

Example 2:

Input: image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, color = 0
Output: [[0,0,0],[0,0,0]]
Explanation: The starting pixel is already colored 0, so no changes are made to the image.

Constraints:

m == image.length
n == image[i].length
1 <= m, n <= 50
0 <= image[i][j], color < 2¹⁶
0 <= sr < m
0 <= sc < n

Solutions

Solution 1

DFS or BFS.

Flood fill, also called seed fill, is a flooding algorithm that determines and alters the area connected to a given node in a multi-dimensional array with some matching attribute. It is used in the “bucket” fill tool of paint programs to fill connected, similarly-colored areas with a different color.

Solution 2: BFS

We first check if the initial pixel’s color is equal to the target color. If it is, we return the original image directly. Otherwise, we can use the breadth-first search method, starting from $(\textit{sr}, \textit{sc})$, to change the color of all eligible pixels to the target color.

Specifically, we define a queue $\textit{q}$ and add the initial pixel $(\textit{sr}, \textit{sc})$ to the queue. Then, we continuously take pixels $(i, j)$ from the queue, change their color to the target color, and add the pixels in the four directions (up, down, left, right) that have the same original color as the initial pixel to the queue. When the queue is empty, we have completed the flood fill.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the 2D array $\textit{image}$, respectively.

class Solution {
    private int[] dirs = {-1, 0, 1, 0, -1};
    private int[][] image;
    private int nc;
    private int oc;

    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        nc = color;
        oc = image[sr][sc];
        this.image = image;
        dfs(sr, sc);
        return image;
    }

    private void dfs(int i, int j) {
        if (i < 0 || i >= image.length || j < 0 || j >= image[0].length || image[i][j] != oc
            || image[i][j] == nc) {
            return;
        }
        image[i][j] = nc;
        for (int k = 0; k < 4; ++k) {
            dfs(i + dirs[k], j + dirs[k + 1]);
        }
    }
}


// Solution 2
class Solution {
    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        if (image[sr][sc] == color) {
            return image;
        }
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {sr, sc});
        int oc = image[sr][sc];
        image[sr][sc] = color;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int i = p[0], j = p[1];
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < image.length && y >= 0 && y < image[0].length
                    && image[x][y] == oc) {
                    q.offer(new int[] {x, y});
                    image[x][y] = color;
                }
            }
        }
        return image;
    }
}

class Solution {
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        int m = image.size(), n = image[0].size();
        int oc = image[sr][sc];
        int dirs[5] = {-1, 0, 1, 0, -1};
        function<void(int, int)> dfs = [&](int i, int j) {
            if (i < 0 || i >= m || j < 0 || j >= n || image[i][j] != oc || image[i][j] == color) {
                return;
            }
            image[i][j] = color;
            for (int k = 0; k < 4; ++k) {
                dfs(i + dirs[k], j + dirs[k + 1]);
            }
        };
        dfs(sr, sc);
        return image;
    }
};


// Solution 2
class Solution {
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        if (image[sr][sc] == color) {
            return image;
        }
        int oc = image[sr][sc];
        image[sr][sc] = color;
        queue<pair<int, int>> q;
        q.push({sr, sc});
        int dirs[5] = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            auto [a, b] = q.front();
            q.pop();
            for (int k = 0; k < 4; ++k) {
                int x = a + dirs[k];
                int y = b + dirs[k + 1];
                if (x >= 0 && x < image.size() && y >= 0 && y < image[0].size() && image[x][y] == oc) {
                    q.push({x, y});
                    image[x][y] = color;
                }
            }
        }
        return image;
    }
};

class Solution:
    def floodFill(
        self, image: List[List[int]], sr: int, sc: int, color: int
    ) -> List[List[int]]:
        def dfs(i, j):
            if (
                not 0 <= i < m
                or not 0 <= j < n
                or image[i][j] != oc
                or image[i][j] == color
            ):
                return
            image[i][j] = color
            for a, b in pairwise(dirs):
                dfs(i + a, j + b)

        dirs = (-1, 0, 1, 0, -1)
        m, n = len(image), len(image[0])
        oc = image[sr][sc]
        dfs(sr, sc)
        return image


# Solution 2
class Solution:
    def floodFill(
        self, image: List[List[int]], sr: int, sc: int, color: int
    ) -> List[List[int]]:
        if image[sr][sc] == color:
            return image
        q = deque([(sr, sc)])
        oc = image[sr][sc]
        image[sr][sc] = color
        dirs = (-1, 0, 1, 0, -1)
        while q:
            i, j = q.popleft()
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < len(image) and 0 <= y < len(image[0]) and image[x][y] == oc:
                    q.append((x, y))
                    image[x][y] = color
        return image

func floodFill(image [][]int, sr int, sc int, color int) [][]int {
	oc := image[sr][sc]
	m, n := len(image), len(image[0])
	dirs := []int{-1, 0, 1, 0, -1}
	var dfs func(i, j int)
	dfs = func(i, j int) {
		if i < 0 || i >= m || j < 0 || j >= n || image[i][j] != oc || image[i][j] == color {
			return
		}
		image[i][j] = color
		for k := 0; k < 4; k++ {
			dfs(i+dirs[k], j+dirs[k+1])
		}
	}
	dfs(sr, sc)
	return image
}


// Solution 2
func floodFill(image [][]int, sr int, sc int, color int) [][]int {
	if image[sr][sc] == color {
		return image
	}
	oc := image[sr][sc]
	q := [][]int{[]int{sr, sc}}
	image[sr][sc] = color
	dirs := []int{-1, 0, 1, 0, -1}
	for len(q) > 0 {
		p := q[0]
		q = q[1:]
		for k := 0; k < 4; k++ {
			x, y := p[0]+dirs[k], p[1]+dirs[k+1]
			if x >= 0 && x < len(image) && y >= 0 && y < len(image[0]) && image[x][y] == oc {
				q = append(q, []int{x, y})
				image[x][y] = color
			}
		}
	}
	return image
}

function floodFill(image: number[][], sr: number, sc: number, newColor: number): number[][] {
    const m = image.length;
    const n = image[0].length;
    const target = image[sr][sc];
    const dfs = (i: number, j: number) => {
        if (
            i < 0 ||
            i === m ||
            j < 0 ||
            j === n ||
            image[i][j] !== target ||
            image[i][j] === newColor
        ) {
            return;
        }
        image[i][j] = newColor;
        dfs(i + 1, j);
        dfs(i - 1, j);
        dfs(i, j + 1);
        dfs(i, j - 1);
    };
    dfs(sr, sc);
    return image;
}


// Solution 2
function floodFill(image: number[][], sr: number, sc: number, color: number): number[][] {
    if (image[sr][sc] === color) {
        return image;
    }

    const oc = image[sr][sc];
    image[sr][sc] = color;

    const q: [number, number][] = [];
    q.push([sr, sc]);

    const dirs = [-1, 0, 1, 0, -1];
    const [m, n] = [image.length, image[0].length];

    while (q.length > 0) {
        const [a, b] = q.shift()!;
        for (let k = 0; k < 4; ++k) {
            const x = a + dirs[k];
            const y = b + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && image[x][y] === oc) {
                q.push([x, y]);
                image[x][y] = color;
            }
        }
    }

    return image;
}

impl Solution {
    fn dfs(image: &mut Vec<Vec<i32>>, sr: i32, sc: i32, new_color: i32, target: i32) {
        if sr < 0 || sr == (image.len() as i32) || sc < 0 || sc == (image[0].len() as i32) {
            return;
        }
        let sr = sr as usize;
        let sc = sc as usize;
        if sr < 0 || image[sr][sc] == new_color || image[sr][sc] != target {
            return;
        }
        image[sr][sc] = new_color;
        let sr = sr as i32;
        let sc = sc as i32;
        Self::dfs(image, sr + 1, sc, new_color, target);
        Self::dfs(image, sr - 1, sc, new_color, target);
        Self::dfs(image, sr, sc + 1, new_color, target);
        Self::dfs(image, sr, sc - 1, new_color, target);
    }
    pub fn flood_fill(image: Vec<Vec<i32>>, sr: i32, sc: i32, new_color: i32) -> Vec<Vec<i32>> {
        let target = image[sr as usize][sc as usize];
        Self::dfs(&mut image, sr, sc, new_color, target);
        image
    }
}


// Solution 2
use std::collections::VecDeque;

impl Solution {
    pub fn flood_fill(mut image: Vec<Vec<i32>>, sr: i32, sc: i32, color: i32) -> Vec<Vec<i32>> {
        let m = image.len();
        let n = image[0].len();
        let (sr, sc) = (sr as usize, sc as usize);

        if image[sr][sc] == color {
            return image;
        }

        let oc = image[sr][sc];
        image[sr][sc] = color;

        let mut q = VecDeque::new();
        q.push_back((sr, sc));

        let dirs = [-1, 0, 1, 0, -1];

        while let Some((i, j)) = q.pop_front() {
            for k in 0..4 {
                let x = i as isize + dirs[k] as isize;
                let y = j as isize + dirs[k + 1] as isize;

                if x >= 0 && x < m as isize && y >= 0 && y < n as isize {
                    let (x, y) = (x as usize, y as usize);
                    if image[x][y] == oc {
                        q.push_back((x, y));
                        image[x][y] = color;
                    }
                }
            }
        }

        image
    }
}

733 - Flood Fill

733. Flood Fill

Description

Solutions

Solution 1

Solution 2: BFS

All Problems

All Solutions

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733. Flood Fill

Description

Solutions

Solution 1

Solution 2: BFS

All Problems

All Solutions